a: \(A=\sqrt{\dfrac{1}{2}}-\dfrac{\sqrt{5}-1}{\sqrt{10}-\sqrt{2}}-\sqrt{3-2\sqrt{2}}\)
\(=\dfrac{1}{\sqrt{2}}-\dfrac{\sqrt{5}-1}{\sqrt{2}\left(\sqrt{5}-1\right)}-\sqrt{\left(\sqrt{2}-1\right)^2}\)
\(=\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{2}}-\left(\sqrt{2}-1\right)\)
\(=-\left(\sqrt{2}-1\right)=-\sqrt{2}+1\)
\(B=\left(3\sqrt{2}-\sqrt{8}\right)\cdot\sqrt{2}\)
\(=\left(3\sqrt{2}-2\sqrt{2}\right)\cdot\sqrt{2}\)
\(=\sqrt{2}\cdot\sqrt{2}=2\)
b:
a: ĐKXĐ: x>=0
\(x-3\sqrt{x}-10=0\)
=>\(\left(\sqrt{x}-5\right)\left(\sqrt{x}+2\right)=0\)
mà \(\sqrt{x}+2>=0\forall x\) thỏa mãn ĐKXĐ
nên \(\sqrt{x}-5=0\)
=>\(\sqrt{x}=5\)
=>x=25(nhận)
b: \(x^2-5x+4=0\)
=>\(\left(x-1\right)\left(x-4\right)=0\)
=>\(\left[{}\begin{matrix}x-1=0\\x-4=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)
